University Physics I - Classical Mechanics, 2019

11.2. SIMPLE HARMONIC MOTION 257
The answer is that there is a very close relationship between simple harmonic motion and circular
motion with constant speed, as Figure 11.3 illustrates: as the point P rotates with constant angular
velocity ω, its projection onto the x axis (the red dot in the figure) performs simple harmonic motion
with angular frequency ω (and amplitude R). (Of course, there is nothing special about the x axis;
the projection on any other axis will also perform simple harmonic motion with the same angular
frequency; for example, the blue dot on the figure.)
If the angular velocity of the particle in Fig. 11.3 is constant, then its “orbital period” (the time
needed to complete one revolution) will be T =2π/ω, and this will also be the period of the
associated harmonic motion (the time it takes for the motion to repeat itself). You can see this
directly from Eq. (11.3): if you increase the time t by 2π/ω, you get the same value of x:
x
(
t + 2π ω
)
= A cos
[
ω
(
t + 2π ω
)]
= A cos(ωt +2π) =A cos(ωt) =x(t) (11.5)
Since the frequency f of an oscillator is equal to 1/T , this gives us the following relationship between
f and ω:
f = 1 T = ω (11.6)
2π
One way to tell whether one is talking about an oscillator’s frequency (f) or its angular frequency
(ω)—apart from the different symbols, of course—is to pay attention to the units. The frequency
f is usually given in hertz, whereas the angular frequency ω is always given in radians per second.
Apart from the factor of 2π, they are, of course, completely equivalent; sometimes one is just
more convenient than the other. On the other hand, the only way to tell whether ω is a harmonic
oscillator’s angular frequency or the angular velocity of something moving in a circle is from the
context. (In this chapter, of course, it will always be the former).
Let us go back now to Eq. 11.3 for our block-on-a-spring system. The derivative with respect to
time will give us the block’s velocity. This is a simple application of the chain rule of calculus:
v(t) = dx = −ωA sin(ωt) (11.7)
dt
Another derivative will then give us the acceleration:
a(t) = dv
dt = −ω2 A cos(ωt) (11.8)
Note that the acceleration is always proportional to the position, only with the opposite sign. The
proportionality constant is ω 2 . Since the force exerted by the spring on the block is F = −kx
(because we are measuring the position from the equilibrium position x 0 ), Newton’s second law,
F = ma, givesus
ma = −kx (11.9)
and you can check for yourself that this will be satisfied if x is given by Eq. (11.3), a is given by
Eq. (11.8), and ω is given by Eq. (11.4).